When Two Differentiation Rules Are Needed
A quotient can have a numerator and a denominator that each require the Chain Rule. In that situation, the Chain Rule does not replace the Quotient Rule: first find the derivative of each part, then use those two derivatives in the quotient formula.
For instance, in \(\frac{(x^2+1)^3}{(x-2)^2}\), the numerator is a power of an expression involving \(x\), and the denominator is also a power of an expression involving \(x\). Differentiating either part requires accounting for its inner function. After that, the quotient rule combines the results.
The condition \(v(x)\ne 0\) matters twice: the original quotient is defined only where its denominator is nonzero, and the quotient-rule formula also divides by the square of that denominator.
A Reliable Order of Work
To keep the two rules distinct, name the entire numerator \(u(x)\) and the entire denominator \(v(x)\). Differentiate each one as a complete function. Only then place \(u\), \(v\), \(u'\), and \(v'\) into the quotient rule.
Set \(u(x)\) equal to the full numerator and \(v(x)\) equal to the full denominator. Note any values excluded because \(v(x)=0\).
Use the Chain Rule wherever an expression is nested inside another function, such as a power.
Form \(u'v-uv'\), in that order, and divide by \(v^2\).
Factoring can make the result easier to read, but it does not make an excluded input part of the domain of the original function.
The subtraction order in Step 3 is essential. The derivative of the numerator is multiplied by the denominator, and then the numerator is multiplied by the derivative of the denominator; the second product is subtracted.
Why the Rules Fit Together
The Quotient Rule can be understood by writing a quotient as a product with a reciprocal. The product rule differentiates that product, while the Chain Rule differentiates the reciprocal of the denominator. This explains both the structure of the formula and why the denominator derivative appears in the subtracted term.
This justification assumes the functions involved are differentiable at the input and \(v\) is nonzero there. In the examples below, the chain-rule derivatives are computed before the quotient-rule substitution, so each factor has a clear source.
Worked Examples
Worked Example: A Powered Polynomial over a Powered Linear Expression
Differentiate \(f(x)=\frac{(x^2+1)^3}{(x-2)^2}\). The denominator is zero at \(x=2\), so the original function is defined for \(x\ne 2\).
Set \(u(x)=(x^2+1)^3\) and \(v(x)=(x-2)^2\). For the numerator, the outer function is a third power and its inner function is \(x^2+1\). Therefore,
For the denominator, the outer function is a square and the inner function is \(x-2\), whose derivative is \(1\). Hence
Now apply the Quotient Rule, keeping the indicated subtraction order:
Factor the common terms \((x^2+1)^2\) and \(x-2\) from the numerator:
For a numerical check, take \(x=0\). The quotient-rule calculation gives \(u(0)=1\), \(v(0)=4\), \(u'(0)=0\), and \(v'(0)=-4\). Thus
The simplified expression also gives \(\frac{1^2(-2)}{(-2)^3}=\frac{-2}{-8}=\frac14\), as required.
Worked Example: Composite Expressions in Both Parts
Differentiate \(g(x)=\frac{(3x-1)^4}{(x^2+1)^2}\). Since \(x^2+1\) is always positive, this quotient is defined for every real \(x\).
Let \(u(x)=(3x-1)^4\) and \(v(x)=(x^2+1)^2\). Apply the Chain Rule separately:
Substitute both derivatives into the quotient rule:
Check at \(x=1\). The original parts have values \(u(1)=2^4=16\) and \(v(1)=2^2=4\). Their derivatives are \(u'(1)=12(2)^3=96\) and \(v'(1)=4(1)(2)=8\). Therefore,
The simplified derivative gives \(\frac{4(2)^3(4)}{(2)^3}=\frac{128}{8}=16\), confirming the result.
Worked Example: A Denominator with a Nonlinear Inner Function
Differentiate \(h(x)=\frac{(2x+1)^2}{(x^2-1)^3}\). The denominator vanishes when \(x^2-1=0\), that is, at \(x=-1\) and \(x=1\). The function is defined only for \(x\ne -1,1\).
Set \(u(x)=(2x+1)^2\) and \(v(x)=(x^2-1)^3\). Their derivatives are
Using the Quotient Rule gives
At \(x=0\), the values are \(u(0)=1\), \(v(0)=(-1)^3=-1\), \(u'(0)=4\), and \(v'(0)=0\). The quotient-rule expression therefore gives
Direct substitution of \(x=0\) into the unsimplified derivative gives the same value: its numerator is \(4(1)(-1)^3-1\cdot0=-4\), and its denominator is \((-1)^6=1\). This check also confirms that the denominator's sign is handled through the powers, rather than being discarded.
A Common Chain-Rule Error
A frequent mistake is to differentiate a power but omit the derivative of its inner expression. In the first example, a student might write \(\frac{d}{dx}(x^2+1)^3=3(x^2+1)^2\). That is incomplete: the inner derivative \(2x\) must also be included.
The error changes the quotient derivative, not merely its appearance. At \(x=1\), the numerator and denominator of \(f\) are \(8\) and \(1\), while the correct derivatives are \(f\)'s numerator derivative \(24\) and denominator derivative \(-2\). The correct quotient-rule calculation gives
If the inner factor \(2x\) is omitted from the numerator derivative, the incorrectly calculated numerator derivative at \(x=1\) is \(3(2)^2=12\). That produces the wrong result
which does not agree with the correct calculation. To avoid this, finish and check each chain-rule derivative before beginning the quotient-rule line.
What to Check Before You Finish
A final expression can look complicated even when the steps are systematic. Before accepting it, check that each part has been differentiated as a whole, that the quotient-rule numerator is \(u'v-uv'\), and that the denominator is \(v^2\). Also retain any exclusions from the original denominator. Cancelling a factor during simplification does not redefine the original function at a point where it was undefined.
These checks are useful because the two rules have different jobs. The Chain Rule supplies the derivative of a composite numerator or denominator. The Quotient Rule then combines those results. Keeping that sequence visible makes it easier to locate an error and easier to verify the final expression.
Check Your Understanding
For each question, identify the domain restriction before differentiating when the denominator may be zero.
- For \(p(x)=\frac{(x^2+3)^2}{(x+1)^3}\), find the derivatives of the numerator and denominator separately before applying the Quotient Rule.
- For \(q(x)=\frac{(4x-1)^3}{(x^2+2)^2}\), which inner derivatives are needed in the two Chain Rule calculations?
- State the Quotient Rule using \(u\) and \(v\), and explain why the condition \(v(x)\ne0\) is required.
- A calculation uses \(u'v+uv'\) in the numerator. Identify the error and write the correct order of terms.
- For \(r(x)=\frac{(x+2)^2}{(x^2-4)^2}\), which inputs are excluded from the domain of the original quotient?
